Do it on a calculator in estimate it's 26.6
Answer:
6
Step-by-step explanation:
54=2x3x3x3
12=2x2x3
common factors are 2 and 3 so 2x3=6
Answer:
![E[R]](https://tex.z-dn.net/?f=E%5BR%5D)
= 99 Ω
= 2.3094 Ω
P(98<R<102) = 0.5696
Step-by-step explanation:
The mean resistance is the average of edge values of interval.
Hence,
The mean resistance, ![E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}](https://tex.z-dn.net/?f=E%5BR%5D%20%3D%20%5Cfrac%7Ba%2Bb%7D%7B2%7D%20%20%3D%20%5Cfrac%7B95%2B103%7D%7B2%7D%20%3D%20%5Cfrac%7B198%7D%7B2%7D)
= 99 Ω
To find the standard deviation of resistance, we need to find variance first.
Hence,
The standard deviation of resistance, 
= 2.3094 Ω
To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.
From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696
Answer:
(−0.103371 ; 0.063371) ;
No ;
( -0.0463642, 0.0063642)
Step-by-step explanation:
Shift 1:
Sample size, n1 = 30
Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm
Shift 2:
Sample size, n2 = 25
Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17
Mean difference ; μ1 - μ2
Zcritical at 95% confidence interval = 1.96
Using the relation :
(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)
(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)
Lower boundary :
-0.02 - 0.0833710 = −0.103371
Upper boundary :
-0.02 + 0.0833710 = 0.063371
(−0.103371 ; 0.063371)
B.)
We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.
C.)
For sample size :
n1 = 300 ; n2 = 250
(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)
Lower boundary :
-0.02 - 0.0263642 = −0.0463642
Upper boundary :
-0.02 + 0.0263642 = 0.0063642
( -0.0463642, 0.0063642)
